6-35
6-52 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area
are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes
are negligible. 3 The device is adiabatic and thus heat transfer is negligible.
Properties From the steam tables (Tables A-4 through 6)
P1
T1
P1 = 10 MPa
T1 = 450qC
V1 = 80 m/s
10 MPa ½ v 1 0.029782 m 3 /kg
¾
450qC ¿ h1 3242.4 kJ/kg
and
P2
x2
10 kPa ½
¾ h2
0.92 ¿
h f x 2 h fg
191.81 0.92 u 2392.1 2392.5 kJ/kg
· STEAM
m = 12 kg/s
Analysis (a) The change in kinetic energy is determined from
'ke
V 22 V12
2
50 m/s
2
(80 m/s) 2
2
§ 1 kJ/kg
¨
¨ 1000 m 2 /s 2
©
·
¸
¸
¹
1.95 kJ/kg
1 m
2 m
. We take the
(b) There is only one inlet and one exit, and thus m
turbine as the system, which is a control volume since mass crosses the
boundary. The energy balance for this steady-flow system can be expressed
in the rate form as
E in E out
'E systemÊ0 (steady)
Rate of net energy transfer
by heat, work, and mass
Rate of change in internal, kinetic,
potential, etc. energies
E in
·
W
P2 = 10 kPa
x2 = 0.92
V2 = 50 m/s
0
E out
# 'pe # 0)
m (h1 V12 / 2) W out m (h2 + V 22 /2) (since Q
W out
§
V 2 V12
m ¨ h2 h1 2
¨
2
©
·
¸
¸
¹
Then the power output of the turbine is determined by substitution to be
W out
(12 kg/s)(2392.5 3242.4 1.95)kJ/kg
10.2 MW
(c) The inlet area of the turbine is determined from the mass flow rate relation,
m
1
v1
o A1
A1V1 
m v 1
V1
(12 kg/s)(0.029782 m 3 /kg )
80 m/s
0.00447 m 2
PROPRIETARY MATERIAL. © 2008 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.